Final answer:
The distance from the point (-2,1,-1) to the plane 3x + 2y + 6z = 5 can be found using the point-to-plane distance formula. By substituting the point coordinates and the plane coefficients into the formula and simplifying, the distance is determined to be √51.
Step-by-step explanation:
To find the distance from the point (-2,1,-1) to the plane 3x + 2y + 6z = 5, we use the point-to-plane distance formula:
D = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²)
Where point P(₁) has coordinates (x₁, y₁, z₁), the plane equation is Ax + By + Cz + D = 0, and A, B, C, D are the coefficients from the equation of the plane. Substituting the values, we get:
D = |3(-2) + 2(1) + 6(-1) + 5| / √(3² + 2² + 6²)
= |-6 + 2 - 6 + 5| / √(9 + 4 + 36)
= |-5| / √49
= 5 / 7
= √(25/49)
= √51/7
Since the result must be a positive distance, we ignore the negative possibility. And the correct option is c) √51.